Amass m = 12 kg is pulled along a horizontal floor, with a coefficient of kinetic friction î¼k = 0.06, for a distance d = 7.8 m. then the mass is continued to be pulled up a frictionless incline that makes an angle î¸ = 25â° with the horizontal. the entire time they massless rope used to pull the block is pulled parallel to the incline at an angle of î¸ = 25â° (thus on the incline it is parallel to the surface) and has a tension t = 20 n. 1) what is the work done by tension before the block gets to the incline

The rope is doing work against friction. The equation for work done by a force is:

W = F×d

W is the work done.

F is the force.

d is the distance covered by the object subjected to the force.

The force of friction is calculated using the following equation:

F = μN

F is the frictional force.

μ is the coefficient of friction between the object and the contact surface.

N is the normal force exerted on the object by the contact surface.

In this situation the normal force is equal to the force of gravity on the object. The normal force is then:

N = mg

m is the mass of the object and g is the acceleration due to gravity of the Earth (9.81m/s²).

Combine all of these equations to get the total work done:

W = μ×m×g×d

Given values:

μ = 0.06

m = 12kg

g = 9.81m/s²

d = 7.8m

Substitute the terms in the equation with the given values and solve for W:

W = 0.06×12×9.81×7.8

the inner core.

good luck: )

klicker